Evaluate $\int\sin^2(\pi x)\cos^5(\pi x)\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(\frac{1}{3}\sin^3(\pi x) - \frac{1}{5}\sin^5(\pi x) + \frac{1}{7}\sin^7(\pi x)\right) + C$ (Choice B) B $\left(\frac{1}{3}\sin^3(\pi x) - \frac{2}{5}\sin^5(\pi x) + \frac{1}{7}\sin^7(\pi x)\right) + C$ (Choice C) C $\frac{1}{\pi}\left(\frac{1}{3}\sin^3(\pi x) - \frac{1}{5}\sin^5(\pi x) + \frac{1}{7}\sin^7(\pi x)\right) + C$ (Choice D) D $\frac{1}{\pi}\left(\frac{1}{3}\sin^3(\pi x) - \frac{2}{5}\sin^5(\pi x) + \frac{1}{7}\sin^7(\pi x)\right) + C$
Answer: We can first use a $~u$ -substitution where $~u=\pi x~$ and $~du =\pi \,dx~$ to simplify the appearance of the integral. $ \int\sin^2(\pi x)\cos^5(\pi x)\,dx~~~~~\Rightarrow~~~~~\dfrac1\pi\int\sin^2u\,\cos^5u\,du$ Here we have an odd power of $~\cos u\,$, so we take out one factor and rewrite the remaining even power of $~\cos u~$ in terms of $~\sin u~$ using the basic identity connecting $~\sin~$ and $~\cos\,$ : $ \cos^2u=1-\sin^2u\,$. $ \begin{aligned}\dfrac1\pi\int\sin^2u\,\cos^5u\,du&=\dfrac1\pi\int\sin^2u\,\big(\cos^2u\big)^2\cos u\,du\\ \\ \\&=\dfrac1\pi\int\sin^2u\,\big(1-\sin^2u\big)^2\cos u\,du\\ \\ \\&=\dfrac{1}{\pi}\int\sin^2u\big(1-2\sin^2u + \sin^4u\big)\cos u\,du\\ \\ \\&=\dfrac{1}{\pi}\int\big(\sin^2u - 2\sin^4u + \sin^6u\big)\cos u\,du\end{aligned}$ We now use a second $~u$ -substitution with $~v=\sin u~$ and $~dv=\cos u\,du\,$. $\begin{aligned}\dfrac{1}{\pi}\int\big(\sin^2u - 2\sin^4u + \sin^6u\big)\cos u\,du&= \dfrac{1}{\pi}\int(v^2 - 2v^4 + v^6)\,dv\\ \\ \\&= \frac{1}{\pi}\left(\frac{1}{3}v^3 - \frac{2}{5}v^5 + \frac{1}{7}v^7\right) + C\end{aligned}$ Now we work our way back from $~v~$ to $~u~$ to $~x\,$. $ \frac{1}{\pi}\left(\frac{1}{3}v^3 - \frac{2}{5}v^5 + \frac{1}{7}v^7\right) + C$ $ \Rightarrow~~~~~ \frac{1}{\pi}\left(\frac{1}{3}\sin^3u - \frac{2}{5}\sin^5 + \frac{1}{7}\sin^7\right) + C$ $ \Rightarrow~~~~~\frac{1}{\pi}\left(\frac{1}{3}\sin^3(\pi x) - \frac{2}{5}\sin^5(\pi x) + \frac{1}{7}\sin^7(\pi x)\right) + C$